MathDB
A 24

Source:

May 25, 2007
floor functionmodular arithmeticDivisibility Theory

Problem Statement

Let p>3p>3 is a prime number and k=2p3k=\lfloor\frac{2p}{3}\rfloor. Prove that (p1)+(p2)++(pk){p \choose 1}+{p \choose 2}+\cdots+{p \choose k} is divisible by p2p^{2}.